被 Hack 的代码（包括我这次修改之前的题解）：

# include <bits/stdc++.h>

# define old_six \
	ios::sync_with_stdio (0);\
	\
	cin.tie (0);\
	\
	cout.tie (0);

# define ffor(i,name) \
	for (auto i = name.begin (); i != name.end (); ++ i)

# define reg register

using namespace std;

typedef long long ll;

typedef pair <int, int> pii;

typedef pair <ll, ll> pll;

int n, a[1005][1005], b[1005][1005], c[1005][1005], dp1[1005][1005], dp2[1005][1005], dp3[1005][1005], sum1, sum2, sum3, max1[1005], max2[1005], max3[1005], maxx, minx = 1e9;

int dfs1 (int x, int y) {

	if (dp1[x][y])
		return dp1[x][y];

	if (x > n - 2)
		return 0;

	return dp1[x][y] = min (dfs1 (x + 1, y) + (a[x + 1][y] != max1[x + 1]), dfs1 (x + 1, y + 1) + (a[x + 1][y + 1] != max1[x + 1]));

}

int dfs2 (int x, int y) {

	if (dp2[x][y])
		return dp2[x][y];

	if (x > n - 2)
		return 0;

	return dp2[x][y] = min (dfs2 (x + 1, y) + (b[x + 1][y] != max2[x + 1]), dfs2 (x + 1, y + 1) + (b[x + 1][y + 1] != max2[x + 1]));

}

int dfs3 (int x, int y) {

	if (dp3[x][y])
		return dp3[x][y];

	if (x > n - 2)
		return 0;

	return dp3[x][y] = min (dfs3 (x + 1, y) + (c[x + 1][y] != max3[x + 1]), dfs3 (x + 1, y + 1) + (c[x + 1][y + 1] != max3[x + 1]));

}

int main () {

	old_six

	cin >> n;

	for (reg int i = 0; i < n; sum1 += max1[i], ++ i)
		for (reg int j = 0; j <= i; ++ j)
			cin >> a[i][j], max1[i] = max (max1[i], a[i][j]);

	for (reg int i = 0; i < n; sum2 += max2[i], ++ i)
		for (reg int j = 0; j <= i; ++ j)
			max2[i] = max (max2[i], b[i][j] = a[n - j - 1][i - j]);
//	for (reg int i = 0; i < n; ++ i, cout << '\n') for (reg int j = 0; j < n; ++ j) cout << b[i][j] << ' ';
	for (reg int i = 0; i < n; sum3 += max3[i], ++ i)
		for (reg int j = 0; j <= i; ++ j)
			max3[i] = max (max3[i], c[i][j] = b[n - j - 1][i - j]);

	maxx = max ({sum1, sum2, sum3});
//	cout << sum1 << ' ' << sum2 << ' ' << sum3 << '\n';
	if (maxx == sum1)
		minx = dfs1 (0, 0);

	if (maxx == sum2)
		minx = min (minx, dfs2 (0, 0) + n);

	if (maxx == sum3)
		minx = min (minx, dfs3 (0, 0) + n * 2);

	cout << maxx << ' ' << minx;

	return 0;

}

Hack 数据生成器：

# include <bits/stdc++.h>
using namespace std;
int main () {
	ios::sync_with_stdio (0);
	cin.tie (0);
	cout.tie (0);
	cout << "1000\n";
	for (int i = 0; i < 1000; ++ i, cout << '\n')
		for (int j = 0; j <= i; ++ j)
			cout << "0 ";
	return 0;
}

Hack 原理：

我使用的是记忆化递归，但是初值赋的是 $0$，而 $a_{i,j}$ 的值可以全部相同，这样可以使 $dp1_{i,j}$ 都变成 $0$，这时候记忆化递归的时间复杂度退化，退成了 $2^n$ 级别，足以导致 TLE